# Homework Help: Force acting on a skier

1. Jun 23, 2011

### PCSL

What forces act on a skier?

I do not want the answer to this question. What I do not understand is why the net force while the skier is accelerating is Fnet=Fthrust-FG. Can someone explain why normal force is not included? Thank you! :)

Sam (75kg) takes off up a 50m high, 10 degree frictionless slope on his jet-powered skis. The skis have a thrust of 200N. He keeps his skis tilted at 10 degrees after becoming airborne.

Last edited: Jun 23, 2011
2. Jun 23, 2011

### cepheid

Staff Emeritus
Re: What forces act on a skier?

Assuming that FG is the component of the weight parallel to the incline, then Fnet is the net force in the direction parallel to the incline. The net force in the direction perpendicular to the incline is zero, because the component of the weight perpendicular to the incline cancels out with the normal force.

3. Jun 23, 2011

### PCSL

Re: What forces act on a skier?

Thank you, but could you clarify my question below?

Last edited: Jun 24, 2011
4. Jun 24, 2011

### PCSL

On second thoughts, I still don't get it. You are correct that FG is the weight component parallel to the incline. However, since weight=mg=gravity, I don't understand why the normal force and gravity/weight wouldn't cancel each other out leaving Fnet=Fthrust=ma. I still don't understand why it is Fnet=Fthrust-mg=ma

5. Jun 24, 2011

### cepheid

Staff Emeritus
If Fg is supposed to be the gravitational force mg (which makes sense), then

Fnet = Fthrust - Fg

is simply not the correct equation. In this situation, as with all such situations, you should draw a free body diagram for the skiier (although in this case it is helpful to include not just the skiier, but the incline as well). What three forces act on him? His weight, the thrust, and the normal force. (Draw all of these on the diagram).

Once you have the diagram, you can see that the weight can be resolved into two components, one which acts parallel to the incline, given by mgsinθ, and one which acts perpendicular to the incline, given by mgcosθ, where θ is the angle of the incline. Then, since there is an acceleration in the direction parallel to the incline, you have that the sum of all forces in that direction is equal to ma:

Fthrust - mgsinθ = ma

Since there is no acceleration in the direction perpendicular to the incline, you can see that the sum of all forces in that direction must be zero:

Fnormal - mgcosθ = 0

Hopefully the answer to your question of why weight doesn't totally cancel out with normal force is now clear from the diagram. Remember that the word normal means "perpendicular" and the normal force is always perpendicular to the contact surface. In contrast, weight is vertical. So they don't even point exactly in the same direction. But if you resolve the weight into parallel and perpendicular components, you find that the perpendicular component of the weight cancels with the normal force. (The parallel component has the effect of hindering the skiier, i.e. of opposing the thrust by trying to pull him down the incline).

6. Jun 24, 2011

### PCSL

That makes sense. Thank you very much. I always draw FBDs but I guess that I just have to work at thinking through problems more.

7. Jun 24, 2011

### WatermelonPig

If you assume he doesn't come off the ground or fall through the ground, then all force acts parallel to the ground and then you use that analysis given by cepheid.

Last edited: Jun 24, 2011